A. $\forall t \in r \left(P\left(t\right)\right)$
B. $\exists t \notin r \left(\neg P\left(t\right)\right)$
solving B,
$\neg\forall t\in r \left(\neg P\left(t\right)\right)$
$\neg\forall t\notin r \left(P\left(t\right)\right)$
$\neg \neg \forall t \in r \left(P\left(t\right)\right)$
$\forall t \in r \left(P\left(t\right)\right)$
which tells us A and B are equivalent, but I'm not sure that negation propagates through $\in$ like the way I did above. Please explain how $\neg$ propagates through $\in$.
First of all the syntax is incorrect: either choose lower-case for sets and upper-case for element, or vise-verse. It is not grammatical to say $x\in t$ or $t\in r$. That is nit-picky, but it helps in rigor.
That said, Your very first step after solving B is wrong, refer to the what was shown above.
Semantic Equivalence:
To say $\forall t \in r(P(t))$ is to say every $t$ in $r$ is $P$ or generally symbolized $P(t)$
Similarly, to say $\exists t \notin r(\lnot P(t))$ is to say there is a $t$ not in $r$ such that it is not $P$ or $P(t)$.
The two are not equivalent: One is saying every $t$ in $r$ is $P$, the other is saying there is a $t$ not in $r$ which is not $P$.
Syntactic Equivalence between $\forall x(x\in Y)$ and $\exists x(x\in Y)$:
Let $Q(x):x\in Y$; therefore, $\forall x(Q(x))\equiv \lnot \exists x(\lnot Q(x))$, but $\lnot Q(x)\equiv\lnot(x\in Y)\equiv x\notin Y$; therefore, $\forall x(Q(x))\equiv\lnot \exists x(x\notin Y)$.
Let $Q(x):x\in Y$; therefore, $\exists x Q(x)\equiv \lnot \forall x(\lnot Q(x))$, but $\lnot Q(x)\equiv \lnot (x\in Y)\equiv x\notin Y$; therefore, $\exists x(x\in Y)\equiv\lnot \forall x(x\notin Y)$
These are the only ways negation can propogate through $\in$
Keeping that in mind: $\forall t \in r(P(t))\equiv \lnot \exists t\in r(\lnot P(t))$ and not $\forall t \in r(P(t))\not \equiv \exists t\notin r(\lnot P(t))$